Given the function #f(x)=(x^2-1)/(x-2)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-1,3] and find the c?

1 Answer
Dec 18, 2016

See below

Explanation:

You determine whether it satisfies the hypotheses by determining whether #f(x) = (x^2-1)/(x-2)# is continuous on the interval #[-1,3]# and differentiable on the interval #(-1,3)#. (Those are the hypotheses of the Mean Value Theorem)

You find the #c# mentioned in the conclusion of the theorem by solving #f'(x) = (f(3)-f(-1))/(3-(-1))# on the interval #(-1,3)#.

#f# is not defined, hence not continuous, at #x = 2#.
Since #2# is in #[-1,3]#, #f# does not satisfy the hypotheses on this interval.

(There is no need to check the second hypothesis.)

To attempt to find #c# try to solve the equation #f'(x) = (f(3)-f(-1))/(3-(-1))#. Discard any solutions outside #(-1,3)#.

The equation has no solution, so there is no #c# that satisfies the conclusion of the Mean Value Theorem.

For those who think attempting to solve the equation was pointless

I suggest you consider

#f(x) = secx# on the interval #[-pi/3,(5pi)/3]#.

Or for a function for which such a #c# exists (though it is a challenge to find) try

#g(x) = x^2/(x^2-1)# on #[-2,4]#.